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Mathematics
Problem
Let
a
a
a
be the first term and
d
d
d
the common difference of an arithmetic progression. Prove that for every positive integer
n
n
n
,
∑
k
=
0
n
−
1
(
a
+
k
d
)
=
n
2
[
2
a
+
(
n
−
1
)
d
]
.
\sum_{k=0}^{n-1}\bigl(a + k d\bigr) = \frac{n}{2}\left[2a + (n-1)d\right].
k
=
0
∑
n
−
1
(
a
+
k
d
)
=
2
n
[
2
a
+
(
n
−
1
)
d
]
.
Provide a detailed proof based only on properties of sigma notation and standard summations.
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